# Class 10 NCERT Solutions- Chapter 5 Arithmetic Progressions – Exercise 5.4

### Question 1. Which term of the AP : 121, 117, 113, . . ., is its first negative term?

### [Hint : Find n for an < 0]

**Solution:**

Given the AP series is 121, 117, 113, . . .,

Here, first term, a = 121 and Common difference, d = 117-121= -4

n^{th}term formula, a_{n}= a+(n −1)dHence,

a

_{n}= 121+(n−1)(-4)= 121-4n+4

=125-4n

To find the first negative term of the series, an < 0 Therefore,

125-4n < 0

125 < 4n

n>125/4

n>31.25

Therefore, the first negative term of the series is

32.^{nd}term (n=32)which will be, a

_{32}=125-4(32) = -3

### Question 2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

**Solution:**

n^{th}term formula, a_{n}= a+(n −1)dThird term, a

_{3}= a+(3 -1)da

_{3}= a + 2dAnd Seventh term, a

_{7}= a+(7-1)da

_{7}= a + 6dAccording to given conditions,

a

_{3}+ a_{7}= 6 …………………………….(i)a

_{3}× a_{7}= 8 ……………………………..(ii)Substituting the values in eqn. (i), we get,

a+2d +a+6d = 6

2a+8d = 6

a+4d=3

a = 3–4d …………………………………(iii)

Now substituting the values in eqn. (ii), we get,

(a+2d)×(a+6d) = 8…………………..(iv)

Putting the value of a from eqn. (iii) in eqn. (iv), we get,

(3–4d +2d)×(3–4d+6d) = 8

(3 –2d)×(3+2d) = 8

3

^{2}– (2d)^{2}= 8 (using the identity,(a+b)(a-b)=a)^{2}-b^{2}9 – 4d

^{2}= 84d

^{2}= 9-8 = 1d = √(1/4)

d = ±1/2

d = 1/2 or -1/2

So now, if

d = 1/2,

a = 3 – 4d = 3 – 4(1/2) = 3 – 2 = 1

and if d = -1/2,

a = 3 – 4d = 3 – 4(-1/2) = 3+2 = 5

Sum of n

^{th}term of AP is:S_{n}= n/2 [2a +(n – 1)d]So,

when a = 1 and d=1/2Then, the sum of first 16 terms are;

S

_{16}= 16/2 [2 +(16-1)1/2] = 8(2+15/2)

S_{16}= 76And

when a = 5 and d= -1/2Then, the sum of first 16 terms are;

S

_{16}= 16/2 [2.5+(16-1)(-1/2)] = 8(5/2)

S_{16}= 20

### Question 3. A ladder has rungs 25 cm apart. (see Figure). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2½ m apart, what is the length of the wood required for the rungs?

### [Hint : Number of rungs = 250/25 + 1]

**Solution:**

As it is given that,

Distance between the rungs of the ladder is 25 cm.

Distance between the top rung and bottom rung of the ladder will be (in cm) = 2 ½ × 100 = 250 cm

Hence, the total number of rungs = 250/25 + 1 = 11

As we can observe here, that, the ladder has rungs in decreasing order from top to bottom. Thus, the rungs are decreasing in an order of AP.

So, According to given condition

First term, a = 45

Last term, l = 25

Number of terms, n = 11

And the length of the wood required for the rungs will be equal to the sum of the terms of AP series formed.Now, as we know, sum of n

^{th}terms is equal to,S_{n}= (n/2)(a+ l)S

_{n}= 11/2(45+25) = 11/2(70) = 385 cmHence,

the length of the wood required for the rungs is 385cm.

### Question 4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

### [Hint : S_{x – 1} = S_{49} – S_{x}]

**Solution:**

According to the given statement,

Row houses are numbered in a row that are in the form of AP, which is as follows:

123…..…..….49Here, according to the given condition,

First term, a = 1

Common difference, d=1

and last term = l

Let us say the number of x

^{th}houses can be represented as:S_{x – 1}= S_{49}– S_{x}Where S

_{x}represents to sum of x^{th}houses.(

Value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it)As, Sum of n

^{th}term of AP,S_{n}= n/2[a+l]Sum of number of houses beyond x house = S

_{x-1}S

_{x-1}= (x-1)/2[1+(x-1)]

S_{x-1}= x(x-1)/2 ………………………………………(i)By the given condition, we can write,

S

_{49}– S_{x}= {49/2[1+(49)}–{x/2(1+x)}

= 25(49) – x(x + 1)/2 ………………………………….(ii)As eqn. (i) and (ii) are equal, So

x(x-1)/2 = 25(49) – x(x-1)/2

x(x-1) = 25(49)

x = ±35

As we know, the number of houses cannot be a negative number.

Hence,

the value of x is 35.

### Question 5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2 m (see Figure). Calculate the total volume of concrete required to build the terrace.

### [Hint : Volume of concrete required to build the first step = 1/4 × 1/2 × 50 m^{3}]

**Solution:**

As we can see from the given figure,

1step is ¼ m high, ½ m wide and 50 m long^{st}

2step is (¼+¼ = ½ m) high, ½ m wide and 50 m long and,^{nd}

3step is (3×¼ = 3/4 m) high, ½ m wide and 50 m long.^{rd}and so on

Hence, we can conclude that the height of step increases by ¼ m each time when width and length is ½ m and 50 m respectively.

So, the height of steps forms a series AP in such a way that;

¼, ½ , ¾, 1, 5/4, ……..

Volume of steps = Volume of Cuboids

= Length × Breadth × HeightNow,

Volume of concrete required to build the first step = ¼ ×½ ×50 = 25/4

Volume of concrete required to build the second step =½ ×½×50 = 25/2

Volume of concrete required to build the second step = ¾ × ½ ×50 = 75/2

Volume of steps = ½ × 50 ×

(¼ + 2/4 + 3/4 + 4/4 + 5/4 + …….) …………………….(1)Now, we can see the height of concrete required to build the steps, are in AP series;

Thus, applying the AP series concept to the height,

First term, a = 1/4

Common difference, d = 2/4-1/4 = 1/4

n = 15

As, the sum of n terms is : S

_{n }= n/2[2a+(n-1)d]S

_{n }= 15/2(2×(1/4 )+(15 -1)1/4)S

_{n}= 15/2 (4)S

_{n}= 30Hence, up solving eqn. (1), we get

Volume of steps = ½ × 50 × S

_{n}= ½ × 50 × 30

= 750 m

^{3}Hence,

the total volume of concrete required to build the terrace is 750 m^{3}.

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